Problem 4

Let

ABC

be an acute-angled triangle with orthocenter

H

, and let

W

be a point on the side

BC

, lying strictly between

B

and

C

. The points

M

and

N

are the feet of the altitudes from

B

and

C

, respectively. Denote by

\omega_1

the circumcircle of

BWN

, and let

X

be the point on

\omega_1

such that

WX

is a diameter of

\omega_1

. Analogously, denote by

\omega_2

the circumcircle of

CWM

, and let

Y

be the point such that

WY

is a diameter of

\omega_2

. Prove that

X, Y

and

H

are collinear.

Step 3 of 5: X and Y lie on the perpendicular to WZ at Z

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\angle WZX = \angle WZY = 90^\circ

Detailed analysis

Since $WX$ is a diameter of $\omega_1$ and $Z\in\omega_1$ , the inscribed angle $\angle WZX=90^\circ$ ; similarly $\angle WZY=90^\circ$ from the diameter $WY$ of $\omega_2$ . Hence $X$ and $Y$ both lie on the line through $Z$ perpendicular to $WZ$ .